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If a stochastic process follows $dX(t) = a(t)X(t)\mathrm dt + \sigma_1(t)X(t)\mathrm dW_1(t)+ \sigma_2(t)X(t)\mathrm dW_2(t)$, does the solution take the form $$ X(t) = X(0) \exp\left(\int_0^t (a(s) - (1/2) \sigma_1(s)^2 - (1/2) \sigma_2(s)^2)\mathrm ds + \int_0^t\sigma_1(s)\mathrm dW_1(s) + \int_0^t\sigma_2(s)\mathrm dW_2(s)\right) $$ where $W_1$ and $W_2$ are 2 Brownian motion (we do not know if they are independent or not).

I cannot use two different $X(t)$ to define the process, as it is assumed that there are 2 stochastic variables $b_1(t)$ and $b_2(t)$ that drive $X(t)$ together. Here, I assume $b_1(t)$ to be a simple stochastic process that follows $$db_1(t) = \sigma_1 dW_1(t)$$ and likewise for $b_2(t)$.

coffee-raid
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  • There seems to be an error in your expression for $X(t)$ - is $\sigma(s)^2$ meant to be $\sigma_2(s)$ ? – Math1000 Mar 12 '24 at 16:02
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    You could try applying Ito's lemma to $\log X_t$ with multidimensional considerations to see what you get. In particular, one should have a term arising from the covariation of $W_1$ and $W_2$ – Jose Avilez Mar 12 '24 at 16:52
  • @Math1000 thank you for pointing that out. I overlooked it while pasting from the solution for a single Brownian motion. – coffee-raid Mar 12 '24 at 17:55
  • @coffee-raid can you specify their dependence of B1 and B2? Do they have constant correlation? – Thomas Kojar Mar 14 '24 at 15:13
  • If they have constant correlation, the answer below will apply again by writing them as sum of some independent BMs. – Thomas Kojar Mar 14 '24 at 15:17
  • @ThomasKojar do you mean the dependence of b1 and b2 on X(t), or the dependence of b1 and b2 on each other? – coffee-raid Mar 14 '24 at 19:30
  • The dependence of X to W1,W2 is clear. I mean the dependence of W1,W2 with each other. The answer will vary a lot depending on their dependence. – Thomas Kojar Mar 14 '24 at 19:34
  • You say " W1 and W2 are 2 Brownian motion (we do not know if they are independent or not)." Can you specify how does W1 depend on W2? – Thomas Kojar Mar 14 '24 at 19:35

1 Answers1

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Indeed, I just noticed that the OP is talking about possibly dependent BMs.

In that case, the OP needs to specify the dependence. If they just have constant correlation $\rho$, we have $W^{1}=B^{1}$ and

$$B^{2}=\sqrt{1-\rho^{2}}W^{2}+\rho B^{1},$$ for two independent BMS $W^1,W^2$. So by rearranging we can just assume that we have two independent BMs but now the coefficients will depend on $\rho$

$\tilde{\sigma}_1=\sigma_1+\sigma_2\rho$ and $\tilde{\sigma}_2=\sigma_2\sqrt{1-\rho^{2}}$.

Using that the two Brownian motions are independent we let

$$Y^{1}_t:=E[X_t| \sigma(W^2)]\text{ and }Y^{2}_t:=E[X_t| \sigma(W^2)].$$

So we have $$X_{t}=Y^{1}_{t}+Y^{2}_{t}-E[X_{t}]$$ (Conditional expectation property for independent sub-sigma algebras). We also get the following two SDEs due to the linearity of the conditional expectation

$$dY^{i}_t=a(t)Y^{i}_{t}dt+\tilde{\sigma}_{i}(t)Y^{i}_{t}dW^{i}_{t},i=1,2$$

and an ODE for $u_{t}:=E[X_{t}]$

$$u_{t}=a(t)u(t)dt\Rightarrow u(t)=u(0)e^{\int^t a(s)ds},$$

where we used that Itō Integral has expectation zero.

Using Solution to General Linear SDE, we also get some concrete solutions $Y^{i}_{t}$

\begin{align*} Y_t^i =& Y_{0}^{i} \exp\left( \int_0^t\left( a(s) - \frac{1}{2}\tilde{\sigma}_{i}^2(s) \right) \mathrm{d}s + \int_0^t \tilde{\sigma}_{i}(s)\mathrm{d}W_s^{i}\right). \end{align*}

Thomas Kojar
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